secants, tangents and properties

7/28/2019 Secants, Tangents and Properties

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Secants, Tangents and Properties 143

20

Secants, Tangents and Properties

8.1 INTRODUCTION

Look at a moving cycle. You will observe that at any instant of time, the wheels of the moving

cycle touch the road at a very limited area, more correctly a point.

If you roll a coin on a smooth surface, say a table or floor, you will find that at any instant

of time, only one point of the coin comes in contact with the surface it is rolled upon.

What do you observe from the above situations ?

Fig. 20.1

If you consider a wheel or a coin as a circle and the touching surface (road or table) as a line,

the above illustrations show that a line touches a circle. In this lesson, we shall study about

the possible contacts that a line and a circle have and try to study their properties.


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20.2 OBJECTIVES

After studying this lesson, the learner will be able to :

z define a secant and a tangent to the circle

z differentiate between a secant and a tangent

z verify and use important results (given in the curriculum) related to tangents and secants

to circles.

20.3 EXPECTED BACKGROUND KNOWLEDGE

z Measurement of angles and line segments

z Drawing circles of given radii

z Drawing lines perpendicular and parallel to given lines

z Knowledge of previous results about lines and angles, congruence and circles.

z Knowledge of Pythagoras Theorem.

20.4 SECANTS AND TANGENTSAN INTRODUCTION

You have read about lines and circles in your earlier lessons. Recall that a circle is the locus

of a point in a plane which moves in such a way that its distance from a fixed point in the

plane always remains constant. The fixed point is called the centre of the circle and the constant

distance is called the radius of the circle. You also know that a line is a collection of points,

extending indefinitely to both sides, whereas a line segment is a portion of a line bounded by

two points.

Fig. 20.2

Now consider the case when a line and a circle co-exist in the same plane. There can be three

distinct possibilities as shown in Fig. 20.2.

You can see that in Fig. 20.2(i), the line XY does not intersect the circle, with centre O. In

other words, we say that the line XY and the circle have no common point. In Fig. 20.2 (ii),

the line XY intersects the circle in two distinct points A and B, and in Fig. 20.2 (iii), the line

XY intersects the circle in only one point and is said to touch the circle at the point P.


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Thus, we can say that in case of intersection of a line and a circle, the following three

possibilities are there :

(i) The line does not intersect the circle at all, i.e., the line lies in the exterior of the circle.

(ii) The line intersects the circle at two distinct points. In that case, a part of the line lies in

the interior of the circle, the two points of intersection lie on the circle and the remainingportion lies in the exterior of the circle.

(iii) The line touches the circle in exactly one point. We therefore define the following :

Tangent

A line which touches a circle at exactly one point is called a tangent line and the

point where it touches the circle is called the point of contact.

Thus, in Fig. 20.2 (iii), XY is a tangent to the circle at P, which is called the point of contact.

Secant

A line which intersects the circle in two distinct points is called a secant line

(usually referred to as a secant).

In Fig. 20.2 (ii), XY is a secant line to the circle and A and B are called the points of intersection

of the line XY and the circle with centre O.

20.5 TANGENT AS A LIMITING CASE OF A SECANT

Consider the secant XY of the circle with centreO, intersecting the circle in the points A and B.

Imagine that one point A, which lies on the circle,

of the secant XY is fixed and the secant rotates

about A, intersecting the circle at B', B'', B''', B''''

as shown in Fig. 20.3 and ultimately attains the

position of the line XAY, when it becomes tangent

to the circle at A.

Thus, we say that

A tangent is the limiting position of a secantwhen the two points of intersection coincide.

20.6 TANGENT AND RADIUS THROUGH THE POINT OF CONTACT

Let XY be a tangent to the circle, with centre O, at the point P. Join OP.

Take points Q, R, S and T on the tangent XY and join OQ, OR, OS and OT.

As Q, R, S and T are points in the exterior of the circle and P is on the circle.

Fig. 20.3


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OP is less than each of OQ, OR, OS and OT.

From our, previous study of Geometry, we know that of

all the segments that can be drawn from a point (not on

the line) to the line, the perpendicular segment is the

shortest :

As OP is the shortest distance from O to the line XY

OP XY

Thus, we can state that

A radius, through the point of contact of tangent to a circle, is perpendicular to

the tangent at that point.

The above result can also be verified by measuring angles OPX and OPY and finding each

of them equal to 90.

20.7 TANGENTS FROM A POINT OUTSIDE THE CIRCLE

Take any point P in the exterior of the circle with centre

O. Draw lines through P. Some of these are shown as PT,

PB, PA, PC, PD and PT' in Fig.20.5.

How many of these touch the circle ? Only two.

Repeat the activity with another point and a circle. You

will again find the same result.

Thus, we can say that

From an external point, two tangents can be

drawn to a circle.

If the point P lies on the circle, can there still be two tangents to the circle from that

point ? You can see that only one tangent can be drawn to the circle in that case. What about

the case when P lies in the interior of the circle ? Note that any line through P in that case

will intersect the circle in two points and hence no tangents can be drawn from an interior

point to the circle.

(A) Now, measure the lengths of PT and PT. You will find that

PT = PT ...(i)

(B) Let us see this logically also.

Consider two s OPT and OPT

OTP = OT'P (Each being a right angle)

OT = OT (Radii of the same circle)

OP = OP (Common)

Fig. 20.5

Fig. 20.4


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OPT OPT

PT = PT ...(ii)

From (A) and (B), we can say that

The lengths of two tangents from an externalpoint are equal.

Also, in Fig. 20.6 as OPT OPT

OPT = OPT


The tangents drawn from an external point to a circle are equally inclined to the

line joining the point to the centre of the circle.

Let us now take some examples to illustrate.

Example 20.1 : In Fig. 20.7, OP = 5 cm and radius of the circle is 3 cm. Find the length of

the tangent PT from P to the circle, with centre O.

Solution : OTP = 90, Let PT = x

In right triangle OTP, we have

OP2 = OT2 + PT2

or 52 = 32 + x2

or x2 = 25 9 = 16

x = 4

i.e. the length of tangent PT = 4 cm

Example 20.2 : In Fig. 20.8, tangents PT and PT are drawn from a point P at a distance of

25 cm from the centre of the circle whose radius is 7 cm. Find the lengths of PT and PT .

Solution : Here OP = 25 cm and OT = 7 cm

We also know that

OTP = 90

PT2 = OP2 OT2

= 625 49 = 576 = (24)2

PT = 24 cm

We also know that

PT = PT

PT = 24 cm

Fig. 20.6

Fig. 20.8

Fig. 20.7


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Example 20.3 : In Fig. 20.9, A, B and C are three exterior points of the circle with centre O.

The tangents AP, BQ and CR are of lengths 3 cm, 4 cm and 3.5 cm. Find the perimeter of

ABC.

Solution : We know that the length of two tangents from an external point to a circle are equal

AP = AR,

BP = BQ,

CQ = CR

AP = AR = 3 cm,

BP = BQ = 4 cm

and CR = CQ = 3.5 cm

AB = AP + PB;

= (3 + 4) cm = 7 cm

BC = BQ + QC;

= (4 + 3.5) cm = 7.5 cm

CA = AR + CR

= (3 + 3.5) cm

= 6.5 cm

Perimeter ofABC = (7 + 7.5 + 6.5) cm = 21 cm

Example 20.4 : In Fig. 20.10, AOB = 50.

Find ABO and OBTSolution : We know that OA XY

OAB = 90

ABO = 180 (OAB + AOB)

= 180 (90 + 50) = 40

We know that OBA = OBT

OBT = 40

ABO = OBT = 40

CHECK YOUR PROGRESS 20.1

1. Fill in the blanks with suitable words :

(i) A tangent is to the radius through the point of contact.

(ii) The lengths of tangents from an external point to a circle are .

(iii) A tangent is the limiting position of a secant when the two coincide.

Fig. 20.9

Fig. 20.10


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(iv) From an external point tangents can be drawn to a circle.

(v) From a point in the interior of the circle, tangent(s) can be drawn to the

circle.

2. In Fig. 20.11, POY = 40, Find the OYP and OYT.

3. In Fig. 20.12, the incircle ofPQR is drawn. If PX = 2.5 cm, RZ = 3.5 cm and perimeter

of PQR = 18 cm , find the length of QY.

Fig. 20.11 Fig. 2012

4. Write an experiment to show that the lengths of tangents from an external point to a circle

are equal.

20.8 INTERSECTING CHORDS INSIDE AND OUTSIDE A CIRCLE.

You have read various results about chords in the previous lesson. We will now verify some

results regarding chords intersecting inside a circle or outside a circle, when produced.

Let us perform the following activity :

Draw a circle with centre O and any radius. Draw two

chords AB and CD intersecting at P inside the circle.

Measure the lengths of the line-segments PD, PC, PA and

PB. Find the products PA PB and PC PD. You will

find that they are equal.

Repeat the above activity with another two circles after

drawing chords intersecting inside. You will again find

that

PA PB = PC PD

Let us now consider the case of chords intersecting outside the circle. Let us perform the

following activity :

Draw a circle of any radius and centre O. Draw two chords BA and DC intersecting each other

outside the circle at P. Measure the lengths of line segments PA, PB, PC and PD. Find the

products PA PB and PC PD.

Fig. 20.13


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You will see that the product PA PB is equal to the

product PC PD, i.e.,

PA PB = PC PD

Repeat this activity with two other circles with chords

intersecting outside the circle. You will again find that

PA PB = PC PD.


If two chords AB and CD of a circle intersect at a point P (inside or outside the

circle), then

PA PB = PC PD.

20.8 INTERSECTING SECANTS AND TANGENTS OF A CIRCLE

To see if there is some relation between the intersecting secantand tangent outside a circle, we conduct the following activity:

Draw a circle of any radius with centre O. From an external point

P, draw a secant PAB and a tangent PT to the circle.

Measure the length of the line-segments PA, PB and PT. Find the

products PA PB and PT PT or PT2. What do you find ?

You will find that

PA PB = PT2

Repeat the above activity with two other circles. You will again find the same result.

Thus, we can say

If PAB is a secant to a circle intersecting the circle at A and B, and PT is a tangent

to the circle at T, then

PA PB = PT2

Let us illustrate these with the help of examples :

Examples 20.5 : In Fig. 20.16, AB and CD are two chords ofa circle intersecting at a point P inside the circle. If PA = 3 cm,

PB = 2 cm, PC = 1.5 cm, find the length of PD.

Solution : It is given that PA = 3 cm, PB = 2 cm and PC = 1.5 cm

Let PD = x

we know that PA PB = PC PD

3 2 = (1.5) x

Fig. 20.14

Fig. 20.15

Fig. 20.16


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x =3 2

15

.=

3 2

1.5= 4

Length of the line-segment PD = 4 cm.

Example 20.6 : In Fig. 20.17, PAB is a secant to the circle from a point P outside the circle.PAB passes through the centre of the circle and PT is a tangent. IfPT = 8 cm and OP = 10 cm, find the radius of the circle, using PA PB = PT2.

Solution : Let x be the radius of the circle

It is given that OP = 10 cm

PA = PO OA = (10 x) cm

and PB = OP + OB = (10 + x) cm

PT = 8 cm

We know that PA PB = PT2

(10 x) (10 + x) = 82

or 100 x2

= 64or x2 = 36 or x = 6

i.e., radius of the circle is 6 cm.

Example 20.7 : In Fig. 20.18, BA and DC are two chords of a circle intersecting each otherat a point P outside the circle. If PA = 4 cm, PB = 10 cm, CD = 3 cm, find PC.

Solution : We are given that PA = 4 cm, PB = 10 cm, CD = 3 cm,

Let PC = x

We know that PA PB = PC PD

or, 4 10 = (x + 3) x

or x2 + 3x 40 = 0

(x + 8) (x 5) = 0

x = 5

PC = 5 cm


1. In Fig. 20.19, if PA = 3 cm, PB = 6 cm, PD = 4 cm, find the length of PC.

2. If in Fig. 20.19, PA = 4 cm, PB = x + 3, PD = 3 cm and PC = x + 5, find the valueof x.

Fig. 20.19 Fig. 20.20 Fig. 20.21

Fig. 20.17

Fig. 20.18


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3. In Fig. 20.20, if PA = 4 cm, PB = 10 cm, PC = 5 cm, find PD

4. In Fig. 20.20, if PC = 4 cm, PD = (x + 5) cm, PA = 5 cm and PB = (x + 2) cm, find

x.

5. In Fig. 20.21, PT = 2 7 cm, OP = 8 cm, find the radius of the circle, if O is the centre

of the circle.

20.9 ANGLES MADE BY A TANGENT AND A CHORD.

Let there be a circle with centre O and let XY be a tangent to the circle at point P. Draw a

chord PQ of the circle through the point P as shown in the Fig. 20.22. Mark a point R on the

major arc and let S be a point on the minor arc .

The segment formed by the major arc and chord PQ is

said to be the alternate segment of QPY and the segment

formed by the minor and chord PQ is said to be the

alternate segment to QPX.

Let us see if there is some relationship between angles in the

alternate segments and the angle between tangents and chord.

Join QR and PR.

Measure PRQ and QPY (See Fig. 20.22)

What do you find ? You will see that PRQ = QPY

Repeat this activity with another circle and same or different radius. You will again find that

QPY = PRQ.

Now measure QPX and QSP. You will again find that these angles are equal.

Thus, we can state that

The angles formed in the alternate segments by a chord through the point of contact

of a tangent to a circle is equal to the angle between the chord and the tangent.

This result is more commonly called as Angle in the Alternate Segment.

Let us now check the converse of the above result.

Draw a circle, with centre O, and draw a chord PQ and let

it form PRQ in alternate segment as shown in Fig. 20.23.

At P, draw QPY = QRP. Extend the line segment PY to

both sides to form line XY. Join OP and measure OPY.

What do you observe ? You will find that OPY = 90 showing thereby that XY is a tangent

to the circle.

Fig. 20.22

Fig. 20.23


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Repeat this activity by taking different circles and you find the same result. Thus, we can state

that

If a line makes with a chord angles which are equal respectively to the angles

formed by the chord in alternate segments, then the line is a tangent to the circle.

Let us now take some examples to illustrate.

Example 20.8 : In Fig. 20.24, XY is tangent to a circle

with centre O. If AOB is a diameter and PAB = 40,

find APX and BPY.

Solution : By the Alternate-segment

theorem, we know that

BPY = BAP

BPY = 40 [Q BAP = 40 (Given)]

Again, APB = 90 [Angle in a semi-circle]

And, BPY + APB + APX = 180 (Angles on a line)

APX = 180 (BPY + APB)

= 180 (40 + 90) = 50.

Example 20.9 : In Fig. 20.25, ABC is an isosceles triangle

with AB = AC and XY is a tangent to the circumcircle of

ABC. Show that XY is parallel to base BC.

Solution : In

ABC, AB = AC 1 = 2

Again XY is tangent to the circle at A.

3 = 2 (Angle in the Alternate segment)

1 = 3

But these are alternate angles

XY || BC


1. Explain with the help of a diagram, the angle formed by a chord in the alternate segment

of a circle.

2. In Fig. 20.26, XY is a tangent to the circle with centre O at a point P. IfOQP = 40,

find the value of a and b.

3. In Fig. 20.27, PT is a tangent to the circle from an external point P. Chord AB of the

circle, when produced meets TP in P. TA and TB are joined and TM is the angle bisector

ofATB.

Fig. 20.24

Fig. 20.25


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Fig. 20.26 Fig. 20.27

IfPAB = 30 and ATB = 60, show that PM = PT.

LET US SUM UP

z A line which intersects the circle in two points is called a secant of the circle.

z A line which touches the circle at a point is called a tangent to the circle.

z A tangent is the limiting position of a secant when the two points of intersection coincide.

z A tangent to a circle is perpendicular to the radius through the point of contact.

z From an external point, two tangents can be drawn to a circle, which are of equal length.

z If two chords AB and CD of a circle intersect at a point P (inside or outside the circle),

then

PA PB = PC PD

z If PAB is a secant to a circle intersecting the circle at A and B, and PT is a tangent to

the circle at T, then

PA PB = PT2

z The angles formed in the alternate segments by a chord through the point of contact of

a tangent to a circle are equal to the angles between the chord and the tangent.

z If a line makes with a chord angles which are respectively equal to the angles formed

by the chord in alternate segments, then the line is a tangent to the circle.

TERMINAL EXERCISE

1. Differentiate between a secant and a tangent to a circle with the help of a figure.

2. Show that a tangent is a line perpendicular to the radius through the point of contact, with

the help of an activity.


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3. In Fig. 20.28, if AC = BC and AB is a diameter

of the circle, find x, y and z.

4. In Fig. 20.29, OT = 7 cm and OP = 25 cm, find

the length of PT. If PT is another tangent to the

circle, find PT and POT.

5. In Fig. 20.30, the perimeter of ABC equals

27 cm. If PA = 4 cm, QB = 5 cm, find the length

of QC.

6. In Fig. 20.30, ifBAC = 70, find BOC.

[Hint : OBC + OCB =12

(ABC + ACB)]

7. In Fig. 20.31, AB and CD are two chords of a

circle intersecting at the interior point P of a

circle. If PA = (x + 3) cm, PB = (x 3) cm,

PD = 3 cm and PC = 5

1

3 cm, find x.

8. In Fig. 20.32, chords BA and DC of the circle,

with centre O, intersect at a point P outside the

circle. If PA = 4 cm and PB = 9 cm, PC = x and

PD = 4x, find the value of x.

Fig. 20.29

Fig. 20.30

Fig. 20.31

Fig. 20.28

Fig. 20.32


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9. In Fig. 20.33, PAB is a secant and PT is a tangent

to the circle from an external point. If PT = x cm,

PA = 4 cm and AB = 5 cm, find x.

10. In Fig. 20.34, O is the centre of the circle and

PBQ = 40. Find

(i) QPY

(ii) POQ

(iii) OPQ Fig. 20.34

Fig. 20.33


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ANSWERS

Check Your Progress 20.1

1. (i) Perpendicular (ii) equal (iii) points of intersection (iv) two (v) no

2. 50, 50

3. 3 cm


1. 4.5 cm 2. 3 cm

3. 8 cm 4. 10 cm

5. 6 cm


2. a = b = 50

Terminal Exercise

3. x = y = z = 45

4. PT = 24 cm; PT = 24 cm, POT = 60

5. QC = 4.5 6. BOC = 125

7. x = 5 8. x = 3

9. x = 6

10. (i) 40 (ii) 80 (iii) 50.

secants, tangents and properties

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